Let $a(x)=14x^6+6x^4-5x^2$, and $b(x)=7x^5+2x^3-3x$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Explanation: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{14x^6+6x^4-5x^2}{7x^5+2x^3-3x}$ : We divide ${7x^5}$ into ${14x^6}$ to get ${2x}$ : $ \hphantom{1567|1444477777} {{2x}}\\ {{{7x^5}+2x^3-3x}}|\overline{{14x^6}+6x^4-5x^2}\\ \hphantom{37...8888........|}\llap{-}\underline{(14x^6+4x^4-6x^2)}\\ \hphantom{37|3....998888890......}{+2x^4+x^2 }\\ $ [What did we do here?] The process stops here because $7x^5+2x^3-3x$ is a polynomial of the fifth degree and $2x^4+x^2$ is a polynomial of the fourth degree. So it follows that ${r(x)}={2x^4+x^2}$, ${q(x)}={2x}$, and $ \dfrac{14x^6+6x^4-5x^2}{7x^5+2x^3-3x}={2x}+\dfrac{{2x^4+x^2}}{7x^5+2x^3-3x}$ To conclude, $q(x)=2x$ $r(x)=2x^4+x^2$